Physics Lesson 16.13.3 - Mutual Induction

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Welcome to our Physics lesson on Mutual Induction, this is the third lesson of our suite of physics lessons covering the topic of Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Mutual Induction

Let's recall the experiment explained in the tutorial 16.7 in which two loops are near each other and a current is induced in one of the loops when turning the switch ON or OFF in the circuit containing the other loop. In addition, the arrow symbol in the resistor means it is a rheostat (variable resistor). As such, the value of current in the circuit can changed not only by turning the switch ON or OFF but also by changing the values of resistance in the rheostat. In this way, we can change the magnitude not only of the induced current but of the corresponding magnetic field as well.

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The process of producing a current through a variable magnetic field is called induction, as explained in precedent tutorials. The induction by which electric current is produced in one coil by changing the magnetic field of the other coil requires the presence of two coils. If one coil is moved away, no current is induced in the other coil due to the long distance. Therefore, the current (and the resulting magnetic field) in one coil produced by this mutual interaction is known as mutual induction.

The mutual induction differs from the self-induction of an inductor, as in the case of inductor the presence of a single solenoid is enough to induce a magnetic field inside it.

When the variable resistor in the figure above is set in a fixed position R, the source produces a steady current I in the circuit. As a result, a magnetic field (and flux) are produced in the coil on the left.

The mutual inductance of the coil 2 (due right) on the coil 1 (due left) is denoted by m₂₁. It is calculated by

M₂₁ = N₂ ∙ Φ₂₁/i₁ (2)

In our figure we have N₁ = N₂ = 1 as both coils have one turn only, but if there are kore turns, we must write their number accordingly.

The above equation has the same form as the equation of inductance for a single coil

L = N ∙ Φ/i

explained in the tutorial 16.9.

We can write the equation (2) as

M₂₁ ∙ i₁ = N₂ ∙ Φ₂₁

If we slightly change the value of resistance R, we obtain a variation of current, so we can write

M₂₁ ∙ di₁/dt = N₂ ∙ dΦ₂₁/dt

From the Faraday's Law, the right side of the above equation represents the emf induced in the coil 2 due to the change in current in the coil 1. Since it opposes the current I1, we obtain

ε₂ = -N₂ ∙ dΦ₂₁/dt = -M₂₁ ∙ di₁/dt

This reasoning can be used for the emf induced in the first coil as well (due to the law of conservation of energy). Therefore, we can write

ε₁ = -N₁ ∙ dΦ₁₂/dt = -M₁₂ ∙ di₂/dt

Experiments show that m₁₂ = m₂₁. Thus, we can write the mutual inductance simply by M. in this way, the two above equations become

ε₁ = -M ∙ di₂/dt

and

ε₂ = -M ∙ di₁/dt

Example 3

A single loop is connected to a 24V battery and a rheostat. The position of rheostat shifts from 20Ω to 8Ω in 2 seconds. This brings an induced current in the second coil, which contains 400 turns. Each coil has an area of 12 cm₂. Calculate:

The mutual inductance if the induced emf in the second coil at the end of process is 120V
The value shown by the ammeter at the end of process
The change of magnetic flux in the second coil
The change in the induced magnetic field in the second coil

Solution 3

Clues:

ε₁ = 24 V
R_i = 20 Ω
R_f = 8 Ω
d_t = 2 s
A₁ = A₂ = A = 12 cm² = 0.0012 m²
B₁ = 400 mT = 0.4 T
N₁ = 1
N₂ = 400
ε₂ = 200 V
a) M = ?
b) i₂ = ?
c) ΔΦ₂ = ?
d) ΔB₂ = ?

First, we calculate the change in current in the first coil due to the change in resistance. We have
i_1(initial) = ε/R₁
= 24 V/20 Ω
= 1.2 A
and
i_1(final) = ε/R₂
= 24 V/8 Ω
= 3 A
Thus,
di₁ = i_1(final) - i_1(initial)
= 3 A - 1.2 A
= 1.8 A
From theory, we know that
ε₂ = -M ∙ di₁/dt
Ignoring the negative sign and substituting the values, we obtain for the mutual inductance:
M = ε₂ ∙ dt/di₁
= (200 V) ∙ (2 s)/(1.8 A)
= 222 H
We use the equation
ε₁ = -M ∙ di₂/dt
to calculate the value shown by the ammeter (which is the value of i₂, i.e. the value of current induced in the second coil), where M is the mutual inductance found at (a). Thus, ignoring again the negative sign, we obtain
di₂ = ε₁ ∙ dt/M
= (24 V) ∙ (2 s)/(222 H)
= 0.216 A
Giving that
ε₂ = -N₂ ∙ dΦ₂₁/dt
we obtain for the change in the magnetic flux in the second coil (the magnitude only):
dΦ₂₁ = ε₂ ∙ dt/N₂
= (200 V) ∙ (2 s)/400
= 1 Wb
Since for a fixed area of loop any change in magnetic flux is due to the change in the magnetic field, we have
∆Φ₂₁ = ∆B₂ ∙ A
∆B₂ = ∆Φ₂₁/A
= 1 Wb/0.0012 m²
= 833.3 T

You have reached the end of Physics lesson 16.13.3 Mutual Induction. There are 3 lessons in this physics tutorial covering Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction, you can access all the lessons from this tutorial below.

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Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.13	Energy Stored in a Magnetic Field. Energy Density of a Magnetic Field. Mutual Induction
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.13.1	Energy Stored in a Magnetic Field
16.13.2	Energy Density of a Magnetic Field
16.13.3	Mutual Induction

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Physics Lesson 16.13.3 - Mutual Induction

Mutual Induction

Example 3

Solution 3

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